Question

如图,在RT△ABC中,∠A=90°,AD⊥BC,D为垂足,∠ABC的平分线交AD,AC于点E,F,请说明：AE=AF

Answer 1

证明：在直角三角形ABF中,∠ABF+∠AFB=90,

在直角三角形EBD中,∠EBD+∠BED=90,

且：∠BED=∠AEF,∠EBD+∠AEF=90,

又：BF为角平分线,则：∠ABF=∠EBD

则可证：∠AEF=∠AFE

则有：AE=AF